VSPER
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Re: VSPER
A molecule that is square planar has 6 regions of electron density, which means that it would have sp3d2 hybridization.
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Re: VSPER
A square planar molecule has 6 regions of electron density, 4 bonded atoms and 2 lone pairs, so it would be sp3d2 hybridized.
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Re: VSPER
The hybridization would be sp3d2. This is due to there being six regions of electron density in a square planar.
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Re: VSPER
VSEPR has to do with regions of electron density, not necessarily just bonds, so it would be sp3d2, to correspond with an octahedral characteristic of electron density.
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Re: VSPER
Hi! So, we look at it from the area of electron density, so since a square planar has 4 bonding areas and 2 lone pairs of electrons, there are 6 electron density regions in total. Thus, having an octahedral electron geometry, the hybridization would be sp3d2.
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Re: VSPER
Hi!
Since it is square planar the hybridization is sp3d2. To find hybridization, I like to figure out the number of regions of electron density. In this case, for a square planar model, the number of regions of electron density is 6. Then, since we know the order is s, p, d, f, and that s can occupy 1, p 3, d 5, and f 7, I start with the lowest and go up. So, s1 would represent 1 number of regions of electron density, sp would be 2, sp2 would be 3, sp3 would be 4, sp3d would be 5, and sp3d2 would be 6, which is the answer in this case.
I hope that this helps!
Since it is square planar the hybridization is sp3d2. To find hybridization, I like to figure out the number of regions of electron density. In this case, for a square planar model, the number of regions of electron density is 6. Then, since we know the order is s, p, d, f, and that s can occupy 1, p 3, d 5, and f 7, I start with the lowest and go up. So, s1 would represent 1 number of regions of electron density, sp would be 2, sp2 would be 3, sp3 would be 4, sp3d would be 5, and sp3d2 would be 6, which is the answer in this case.
I hope that this helps!
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